Question Number 69795 by mathmax by abdo last updated on 27/Sep/19
![let p(x)=(x+in)^n −n^n with n integr natural 1) find the roots of p(x) 2)factorize p(x) inside C[x] 3) decompose the fraction F(x)=(1/(p(x)))](https://www.tinkutara.com/question/Q69795.png)
$${let}\:{p}\left({x}\right)=\left({x}+{in}\right)^{{n}} −{n}^{{n}} \:{with}\:{n}\:{integr}\:{natural} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){factorize}\:{p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right)\:{decompose}\:{the}\:{fraction}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{p}\left({x}\right)} \\ $$
Commented by mathmax by abdo last updated on 29/Sep/19
![1)p(x)=0 ⇔(x+in)^n =n^n ⇔(((x+in)^n )/n^n ) =1 ⇔(((x+in)/n))^n =1 ⇔((x/n)+i)^n =1 let z =(x/n)+i (e)⇒z^n =1 ⇒z^n =e^(i(2kπ)) ⇒ z_k =e^((i2kπ)/n) with k∈[[0,n−1]] so we get nz_k =x_k +ni ⇒ x_k =n(z_k −i) =n(e^((i2kπ)/n) −i) with0≤k≤n−1 2) p(x) =aΠ_(k=0) ^(n−1) (x−x_k ) =aΠ_(k=0) ^(n−1) (x+ni−ne^((i2kπ)/n) ) we have p(x)=Σ_(k=0) ^n C_n ^k x^k (in)^(n−k) −n^n k=n ⇒a =C_n ^n (in)^0 =1 ⇒p(x)=Π_(k=0) ^(n−1) (x+ni−ne^((i2kπ)/n) ) 3) F(x)=(1/(p(x))) =(1/(Π_(k=0) ^(n−1) (x−x_k ))) =Σ_(k=0) ^(n−1) (λ_k /(x−x_k )) λ_k =(1/(p^′ (x_k )))](https://www.tinkutara.com/question/Q69960.png)
$$\left.\mathrm{1}\right){p}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\left({x}+{in}\right)^{{n}} ={n}^{{n}} \:\Leftrightarrow\frac{\left({x}+{in}\right)^{{n}} }{{n}^{{n}} }\:=\mathrm{1}\:\Leftrightarrow\left(\frac{{x}+{in}}{{n}}\right)^{{n}} =\mathrm{1}\: \\ $$$$\Leftrightarrow\left(\frac{{x}}{{n}}+{i}\right)^{{n}} =\mathrm{1}\:\:{let}\:{z}\:=\frac{{x}}{{n}}+{i}\:\:\:\:\left({e}\right)\Rightarrow{z}^{{n}} =\mathrm{1}\:\Rightarrow{z}^{{n}} ={e}^{{i}\left(\mathrm{2}{k}\pi\right)} \:\Rightarrow \\ $$$${z}_{{k}} ={e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \:\:{with}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:\:\:{so}\:{we}\:{get}\:{nz}_{{k}} ={x}_{{k}} +{ni}\:\Rightarrow \\ $$$${x}_{{k}} ={n}\left({z}_{{k}} −{i}\right)\:={n}\left({e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} −{i}\right)\:\:\:\:{with}\mathrm{0}\leqslant{k}\leqslant{n}−\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:\:{p}\left({x}\right)\:={a}\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{x}_{{k}} \right)\:={a}\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}+{ni}−{ne}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \right) \\ $$$${we}\:{have}\:{p}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \left({in}\right)^{{n}−{k}} \:−{n}^{{n}} \\ $$$${k}={n}\:\Rightarrow{a}\:={C}_{{n}} ^{{n}} \:\left({in}\right)^{\mathrm{0}} =\mathrm{1}\:\Rightarrow{p}\left({x}\right)=\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}+{ni}−{ne}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \right) \\ $$$$\left.\mathrm{3}\right)\:{F}\left({x}\right)=\frac{\mathrm{1}}{{p}\left({x}\right)}\:=\frac{\mathrm{1}}{\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{x}_{{k}} \right)}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\lambda_{{k}} }{{x}−{x}_{{k}} } \\ $$$$\lambda_{{k}} =\frac{\mathrm{1}}{{p}^{'} \left({x}_{{k}} \right)} \\ $$